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Problem 296

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claims/: The 1 claim page of Problem 296, one per claimant's result; the problem's standing derives from them.


Statement. Let N≥1N\geq 1 and let k(N)k(N) be maximal such that there are kk disjoint A1,…,Ak⊆{1,…,N}A_1,\ldots,A_k\subseteq \{1,\ldots,N\} with $\sum_{n\in A_i}\frac{1}{n}=1$ for all ii. Estimate k(N)k(N). Is it true that $k(N)=o(\log N)$?

Formulation. The site's wording on 2026-09-17 (the page shows no last-edited date). The AiA_i are pairwise disjoint and each has reciprocal sum exactly one; k(N)≥1k(N)\ge1 because of {1}\{1\}, and disjointness gives k(N)≤∑n≤N1/nk(N)\le\sum_{n\le N}1/n. The closing question is Erdős and Graham's guess ("No doubt k=o(log⁡n)k=o(\log n) but we have not proved this", monograph printed p. 36).

Status. PROVED (LEAN) is the site's label, which attaches to the estimate: k(N)=(1−o(1))log⁡Nk(N)=(1-o(1))\log N, by Bloom's Theorem 3 (J. Eur. Math. Soc. 27 (2025)) combined with a greedy removal argument, an observation the site credits to Hunter and Sawhney and which is written out below. This is the accepted claim Hunter and Sawhney, recorded with the value answered: it determines k(N)k(N) to first order and answers the closing question in the negative, since k(N)/log⁡N→1k(N)/\log N\to1 rules out k(N)=o(log⁡N)k(N)=o(\log N). The Lean suffix is a catalog label qualified under Existing formalization below; the Lean files are formalization links on the claim page, and no local kernel credit is claimed.

Source. erdosproblems.com/296, accessed 2026-09-17: the problem page (PROVED (LEAN)), its discussion thread (two comments, of 22 April and 26 May 2026) and its empty proof-claim tab. The site cites [ErGr80] as the problem's source and [Bl21] in its commentary. Cite as: T. F. Bloom, Erdős Problem #296, https://www.erdosproblems.com/296, accessed 2026-09-17.

References.

  • [ErGr80] Erdős, P. and Graham, R. L., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique 28, Université de Genève (1980), printed p. 36. Library home: erdos_1980_old_new_problems_results_combinatorial_number_theory.
  • [Bl21] Bloom, T. F., On a density conjecture about unit fractions, with an appendix co-written by Bloom and B. Mehta. arXiv:2112.03726 (2021), v2 (12 October 2023); J. Eur. Math. Soc. 27 (2025), no. 11, 4563--4589, DOI 10.4171/JEMS/1456. Theorem 3 (Theorem 1.3 in the published version). Library home: bloom_2021_density_conjecture_about_unit_fractions.
  • [LiSa24] Liu, Y. P. and Sawhney, M., On further questions regarding unit fractions. arXiv:2404.07113v1 (2024); Int. Math. Res. Not. 2026, no. 2, rnaf382. Theorem 1.1, which sharpens the error term below. Library home: liu_2024_further_questions_regarding_unit_fractions.

Formalization. Statement in ErdosProblems/296.lean of formal-conjectures, as of 2026-09-17, with an external proof tag; the thread links a conditional and an unconditional Lean 4 file. None was built by this corpus. See Existing formalization.

Current assessment

The question. On 2026-09-17 the site asks for the order of k(N)k(N) and whether k(N)=o(log⁡N)k(N)=o(\log N) and shows PROVED (LEAN). Its commentary credits Hunter and Sawhney with noticing that Bloom's Theorem 3, applied repeatedly by a greedy removal, gives k(N)=(1−o(1))log⁡Nk(N)=(1-o(1))\log N, and raises the wider question of how many pairwise disjoint subsets of {1,…,N}\{1,\ldots,N\} can share one reciprocal sum, for which it cites a sunflower argument giving at least Nexp⁡(−O(log⁡N))N\exp(-O(\sqrt{\log N})) sets. The two comments in the thread concern Lean formalizations and are recorded under Existing formalization. The proof-claim tab is empty. The community database record (teorth/erdosproblems) lists the problem as proved (Lean) with the statement formalized and no formal-proof URL, as of its last update of 22 April 2026.

Status support. The status-defining source is Bloom's Theorem 3 (arXiv:2112.03726v2, p. 2; Theorem 1.3 in the published version, J. Eur. Math. Soc. 27 (2025), 4563--4589, accepted 11 October 2023 and online 11 July 2024 per the publisher's record of 2026-09-17): there is an absolute C>0C>0 such that, for NN sufficiently large, every A⊆{1,…,N}A\subseteq\{1,\ldots,N\} with reciprocal sum at least T(N)=Clog⁡Nlog⁡log⁡log⁡N/log⁡log⁡NT(N)=C\log N\log\log\log N/\log\log N contains a subset with reciprocal sum one. The library holds a complete rewritten proof of Theorem 3 (not independently reviewed) through the explicit variant of its Proposition 1 used by the existing formalization. The passage from Theorem 3 to the estimate of k(N)k(N) is the elementary greedy argument in the next section, written for this page and not taken from a source; it needs nothing beyond Theorem 3 and the size of the harmonic sum. The site attributes the observation to Hunter and Sawhney.

The answer to the closing question is no. Since k(N)/log⁡N→1k(N)/\log N\to1, k(N)k(N) is not o(log⁡N)o(\log N). The site's label PROVED belongs to the estimate; the monograph's guess is refuted by the same estimate.

Search scope (2026-09-17 UTC). The site's problem, discussion and proof-claim pages on 2026-09-17; the community database record; the formal-conjectures file, the external Lean file it tags and the two files the thread links; the arXiv listing for 2112.03726 (v1 7 December 2021, v2 12 October 2023); the EMS article record; the Semantic Scholar citing-paper records for Bloom's paper (nine records; the 2025 and 2026 items concern approximate reciprocal subsums, partitions with prescribed reciprocal sums and the count of unit-sum subsets, none this problem); the arXiv API listing of the sixty most recent abstracts mentioning unit or Egyptian fractions (to 7 September 2026); and two general web searches. Not searched: MathSciNet, zbMATH, full-text search engines for scholarly literature, X. Nothing found refines the error term below or concerns the equal-sums variant; this is a bounded negative finding.

Remaining gaps. The equal-sums variant (the site's "more generally") is recorded only as the site's and the monograph's statement; the sunflower argument is not compiled. The greedy deduction has no independent review. The Lean files are not built or audited by this corpus.

The estimate and its deduction

Write HN=∑n≤N1/nH_N=\sum_{n\le N}1/n, so that log⁡N<HN≤1+log⁡N\log N<H_N\le1+\log N for N≥1N\ge1.

Upper bound. If A1,…,Ak⊆{1,…,N}A_1,\ldots,A_k\subseteq\{1,\ldots,N\} are pairwise disjoint with reciprocal sum one each, then k=∑i∑n∈Ai1/n≤HNk=\sum_{i}\sum_{n\in A_i}1/n\le H_N. Hence k(N)≤1+log⁡Nk(N)\le1+\log N.

Lower bound. Let CC and N0N_0 be as in Theorem 3 and let N≥N0N\ge N_0 with HN≥T(N)H_N\ge T(N). Put B0={1,…,N}B_0=\{1,\ldots,N\}. If Bj⊆{1,…,N}B_j\subseteq\{1,\ldots,N\} has reciprocal sum at least T(N)T(N), Theorem 3 gives Sj+1⊆BjS_{j+1}\subseteq B_j with reciprocal sum one; put Bj+1=Bj∖Sj+1B_{j+1}=B_j\setminus S_{j+1}, whose reciprocal sum is exactly one less. The sets S1,S2,…S_1,S_2,\ldots obtained are pairwise disjoint subsets of {1,…,N}\{1,\ldots,N\} with reciprocal sum one. After jj removals the remaining reciprocal sum is HN−jH_N-j, so a further removal is possible as long as HN−j≥T(N)H_N-j\ge T(N), and the number of sets produced is at least ⌊HN−T(N)⌋+1>HN−T(N)>log⁡N−T(N)\lfloor H_N-T(N)\rfloor+1>H_N-T(N)>\log N-T(N). Therefore

log⁡N−C log⁡log⁡log⁡Nlog⁡log⁡N log⁡N<k(N)≤1+log⁡N(N≥N0),\log N-C\,\frac{\log\log\log N}{\log\log N}\,\log N<k(N)\le1+\log N \qquad(N\ge N_0),

which is k(N)=(1−o(1))log⁡Nk(N)=(1-o(1))\log N. Replacing Theorem 3 by Liu and Sawhney's Theorem 1.1 gives, for every fixed ε>0\varepsilon>0 and NN large in terms of ε\varepsilon, the sharper lower bound k(N)>log⁡N−(log⁡N)4/5+εk(N)>\log N-(\log N)^{4/5+\varepsilon}. The argument is the greedy removal the site's commentary refers to; it is elementary and carries no independent review.

History and the equal-sums variant

Printed p. 36 of the Erdős–Graham monograph asks "How many disjoint sets Si∈XS_i\in\mathscr X, 1≤i≤k1\le i\le k, can we find so that Si⊆{1,2,…,n}S_i\subseteq\{1,2,\ldots,n\}? No doubt k=o(log⁡n)k=o(\log n) but we have not proved this", where X\mathscr X is the family of finite sets of integers with reciprocal sum one, and continues: "More generally, how many disjoint sets Ti⊆{1,2,…,n}T_i\subseteq\{1,2,\ldots,n\} are there so that all the sums ∑t∈Ti1/t\sum_{t\in T_i}1/t are equal. [sic] By using strong Δ\Delta-systems [Er-Ra (60)], it can be shown that there are at least n/eclog⁡nn/e^{c\sqrt{\log n}} such TiT_i. Is this the right order of magnitude?" The site's commentary repeats the second question with the same bound. The sunflower argument is not given in either place and no reconstruction of it is compiled; the equal-sums question is a variant that the estimate of k(N)k(N) does not answer.

Existing formalization

The formal-conjectures file ErdosProblems/296.lean, at the revision the Formalization link above pins (2026-09-17), defines recipSum and HasDisjointUnitDecomps N k (there are kk pairwise disjoint subsets of Finset.Icc 1 N with reciprocal sum one) and declares

erdos_296 : (∀ N k : ℕ, HasDisjointUnitDecomps N k → (k : ℚ) ≤ recipSum (Finset.Icc 1 N)) ∧ (∀ ε : ℝ, 0 < ε → ε < 1 → ∀ᶠ N : ℕ in atTop, HasDisjointUnitDecomps N ⌊(1 - ε) * Real.log N⌋₊)

under category research solved with proof sorry and the attribute formal_proof using lean4 at the file src/v4.29.1/ErdosProblems/Erdos296.lean of the collection plby/lean-proofs. This is the two-sided estimate above, not the yes-or-no question. The external file (branch main, as of 2026-09-17, linked from the claim page at that revision) names Bloom, Hunter and Sawhney as informal authors and "Aristotle" and John Jennings as formal authors, with a copyright line "John Jennings, Aristotle (Harmonic)"; it imports UnitFractions.ErdosProblems and Mathlib, proves erdos296_upper_bound (the first conjunct), bloom_quantitative (Theorem 3, obtained from UnitFractions.unit_fractions_upper_log_density), erdos296_answer (the second conjunct) and finally

erdos296 : ∃ c : ℝ, c > 0 ∧ ∀ᶠ N : ℕ in atTop, HasDisjointUnitDecomps N ⌊c * Real.log N⌋₊

with c=1/2c=1/2, without sorry, and ends with a comment recording #print axioms erdos296 as propext, Classical.choice, Quot.sound. Aristotle is an automated prover; its authorship is recorded here as provenance, and no independent check of the file is claimed.

The thread's first comment (22 April 2026) announces a conditional formalization produced by Aristotle, a single-file gist run on the Lean web editor, whose one non-standard axiom unit_fractions_upper_log_density corresponds to Theorem 3 and is said to be semantically identical to the proven Lean 3 theorem at line 2042 of src/final_results.lean in the Bloom–Mehta repository (b-mehta/unit-fractions, master branch on 2026-09-17). The second comment (26 May 2026) links an unconditional standalone file, problems/296/Erdos296.lean in the repository Jayyhk/erdos-lean (main branch on 2026-09-17, linked from the claim page at that revision; 16,481 lines; imports only Mathlib), which says it vendors the Lean 4 port of the Bloom–Mehta formalization from plby/unit-fractions and proves

erdos_296 : ∀ ε : ℝ, 0 < ε → ε < 1 → ∀ᶠ N : ℕ in atTop, HasDisjointUnitDecomps N ⌊(1 - ε) * Real.log N⌋₊

with a comment recording the same three axioms. None of the three files was built, audited or kernel-checked by this corpus; the site's Lean suffix is a catalog label, and the community database records no formal-proof URL. The three files are formalization links on the claim page, which lists no formalized evidence.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.